The nozzle in Fig. 13–39a is connected to a large reservoir where the absolute air pressure is 350 kPa. Determine the range of outside backpressures that cause a shock wave to form within the nozzle and just outside of it.

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Fluid Description. We assume steady flow through the nozzle.

Analysis. First we will establish the backpressures that produce subsonic and supersonic isentropic flow through the nozzle, curves 3 and 4 in Fig. 13–39b. The area ratio between the exit and the throat of the nozzle (divergent section) is [latex] A / A^*=\pi(0.125 m )^2 / \pi(0.0625 m )^2=4 [/latex]. This ratio gives two values for the Mach number at the exit using Eq. 13–36 or Table B–1 (k = 1.4). Choosing the isentropic subsonic flow within the divergent section (curve 3) for [latex] A / A^*=4 ext {, we get } M \approx 0.1465<1 [/latex] and [latex] p / p_0 [/latex] = 0.9851. Since there is no shock wave, the stagnation pressure throughout the flow is 350 kPa. Then, at the exit, [latex] p_3 [/latex] = 0.9851(350 kPa) = 345 kPa. In other words, this backpressure will cause M = 1 at the throat and subsonic isentropic flow of M = 0.1465 at the exit.

For isentropic supersonic flow within the divergent region (curve 4), from Eq. 13–36, for A/A* = 4, we get M = 2.9402 > 1 and [latex] p / p_0 [/latex] = 0.02979.
Thus, at the exit, [latex] p_4 [/latex] = 0.02979(350 kPa) =10.43 kPa = 10.4 kPa.
This pressure is lower than the previous value, since it must produce the required supersonic flow at the exit of M = 2.9402. The two solutions are for isentropic flow with backpressures that produce no shock within the nozzle; however, for both conditions the nozzle is choked since M = 1 at the throat.
If a standing shock is formed at the exit of the nozzle, Fig. 13–39b, curve 6, then this will occur when the pressure in the nozzle at the exit, to the left or behind the shock, is 10.43 kPa. To find the backpressure in front of the shock, that is, just outside the exit plane, we must use Eq. 13–76 or Table B–4 with M = 2.9402, in which case [latex] p_6 / p_4 [/latex] = 9.9187, so that [latex] p_6=9.9187, p_4=9.9187(10.43 kPa )=103 kPa [/latex].
Thus, a normal shock is created within the divergent section (such as curve 5, which is between curves 3 and 6) when the backpressure is in the following range:

[latex] \frac{p_2}{p_1}=\frac{2 k}{k+1} M _1^2-\frac{k-1}{k+1} [/latex] (13-76)

[latex] \frac{A}{A^*}=\frac{1}{ M }\left[\frac{1+\frac{1}{2}(k-1) M ^2}{\frac{1}{2}(k+1)} ight] \frac{k+1}{2(k-1)} [/latex] (13-36)

[latex] 103 kPa

For compression waves to occur at the exit, curve 7, the backpressure (between curves 6 and 4) should be in the range

[latex] 10.4 kPa

Finally, expansion waves will form if the backpressure is anywhere below curve 4 (as in curve 8), that is,

[latex] p_b<10.4 kPa [/latex]

Question 13.18: The nozzle in Fig. 13–39a is connected to a large reservoir ...

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